Tag: Mean Value Theorem

Prove the integral from 0 to x of sin t /(t² + 1) is positive for positive x

by RoRi

October 30, 2015

Prove that for all $x \geq 0$ we have

$\int_0^x \frac{\sin t}{t+1} \, dt \geq 0.$

Before proceeding with the proof, we recall the second mean-value theorem for integrals (Theorem 5.5 on p. 219 of Apostol). For a continuous function $g$ on the interval $[a,b]$ if $f$ has a continuous derivative which never changes sign on the interval $[a,b]$ then there exists a $c \in [a,b]$ such that

$\int_a^b f(x) g(x) \, dx = f(a) \int_a^c g(x) \, dx + f(b) \int_c^b g(x) \, dx.$

Proof. Now, we want to apply the mean-value theorem above with $g(t) = \sin t$ and $f(t) = \frac{1}{t+1}$ . Since $\sin t$ is continuous everywhere $\mathbb{R}$ , it is continuous on any interval $[0,x]$ . Then,

$f'(t) = \frac{-1}{(1+t)^2}$

is continuous for all $t \neq -1$ (so, in particular, for all $t \geq 0$ ). Furthermore, since $(1+t)^2 > 0$ for all $t \geq 0$ we have that $f'(t) < 0$ for all $t \geq 0$ . Thus, $f'(t)$ is continuous and never changes sign in any interval $[0,x]$ . Therefore, we can apply the mean-value theorem to conclude there exists a $c \in [0,x]$ for any $x \geq 0$ such that

$\begin{align*} \int_0^x \left(\frac{1}{1+t}\right) \sin t \, dt &= \frac{1}{1+0} \int_0^c \sin t \, dt + \frac{1}{1+1} \int_c^x \sin t \, dt \\ &= (- \cos t) \Big \rvert_0^c + \frac{1}{2} (-\cos t) \Big \rvert_c^x \\ &= 1 - \cos c + \frac{1}{2} \cos c - \frac{1}{2} \cos x \\ &= 1 - \frac{1}{2} ( \cos c + \cos x). \end{align*}$

But since $\cos x \leq 1$ for all $x$ we know $\cos c + \cos x \leq 2$ for any $c$ and $x$ . Hence,

$\int_0^x \frac{\sin t}{1+t} \, dt = 1 - \frac{1}{2}(\cos c + \cos x) \geq 0. \qquad \blacksquare$

Application of the mean-value theorem for integrals

by RoRi

October 23, 2015

Let $\varphi$ be a function with second derivative $\varphi''$ continuous and nonzero on an interval $[a,b]$ . Furthermore, let $m > 0$ be a constant such that
$\varphi'(t) \geq m \qquad \text{for all } t \in [a,b].$

Use the second mean-value theorem for integrals (Theorem 5.5 in Apostol) to prove the inequality

$\left| \int_a^b \sin \varphi(t) \, dt \right| \leq \frac{4}{m}.$
If $a > 0$ prove that
$\left| \int_a^x \sin (t^2) \, dt \right| \leq \frac{2}{a} \qquad \text{for } x > a.$

Proof. Since we have the assumption that $\varphi'(t) \geq m > 0$ for all $t \in [a,b]$ we may divide by $\varphi'(t)$ , to obtain
$\left| \int_a^b \sin \varphi(t) \, dt \right| = \left| \int_a^b \frac{\sin \varphi(t)}{\varphi'(t)} \cdot \varphi'(t) \, dt \right|.$

Then, to apply the second mean-value theorem for integrals (Theorem 5.5 of Apostol) we define functions

$f(t) = \frac{1}{\varphi'(t)} \qquad \text{and} \qquad g(t) = \varphi'(t) \sin \varphi(t).$

The function $g$ is continuous since $\sin \varphi(t)$ is a composition of continuous functions (we know $\varphi(t)$ is continuous since it is differentiable) and $\varphi'(t)$ is continuous (again, it is differentiable since $\varphi''(t)$ exists and is continuous by assumption). Then the product of continuous function is also continuous, which establishes that $g$ is continuous. We also know that $f$ meets the conditions of the theorem since it has derivative given by

$f'(t) = - \frac{\varphi''(t)}{(\varphi'(t))^2}.$

This derivative is continuous since $\varphi''(t)$ and $\varphi'(t)$ are continuous and $\varphi'(t)$ is nonzero. Furthermore, this derivative does not change since on $[a,b]$ since $\varphi''(t)$ is nonzero on $[a,b]$ (and by Bolzano’s theorem we know that a continuous function that changes sign must have a zero). Therefore, we can apply the second mean-value theorem:

$\begin{align*} \left| \int_a^b \frac{\varphi'(t) \sin \varphi(t)}{\varphi'(t)} \, dt \right| &= \left| \frac{1}{\varphi'(a)} \int_a^c \varphi'(t) \sin \varphi(t) \, dt + \frac{1}{\varphi'(b)} \int_c^b \varphi'(t) \sin \varphi(t) \, dt \right| \\ &\leq \left| \frac{1}{m} \int_a^c \varphi'(t) \sin \varphi(t) \, dt + \frac{1}{m} \int_c^b \varphi'(t) \sin \varphi(t) \, dt \right| \\ &\leq \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_a^c + \left(-\frac{1}{m} \right) \cos \varphi(t) \Big \rvert_c^b \right| \\ & \leq \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_a^c \right| + \left| -\frac{1}{m} \cos \varphi(t) \Big \rvert_c^b \right| \\ \intertext{(by the Triangle Inequality)} &\leq \left| -\frac{2}{m} \right| + \left| -\frac{2}{m} \right| \\ & = \frac{2}{m} + \frac{2}{m} \\ &= \frac{4}{m}. \qquad \blacksquare \end{align*}$
Proof. Using part (a), we take $\varphi(t) = t^2$ , giving us
$\varphi(t) = t^2 \quad \implies \quad \varphi'(t) = 2t \quad \implies \quad \varphi''(t) = 2.$

Thus, $\varphi''(t)$ is continuous and never changes sign. Furthermore, $\varphi' \geq m = 2a$ (where $a$ is a given constant) and $2a > 0$ since $a> 0$ . Thus,

$\left| \int_a^x \sin (t^2) \, dt \right| \leq \frac{4}{2a} = \frac{2}{a} \qquad \text{for all } x > a. \qquad \blacksquare$

Find a function with continuous second derivative satisfying given conditions

by RoRi

October 16, 2015

In each of the following cases find a function $f$ with continuous second derivative $f''$ satisfying the given conditions.

$f''(x) > 0$ for all $x$ , $f'(0) = 1$ , and $f'(1) = 0$ .
$f''(x) > 0$ for all $x$ , $f'(0) = 1$ , and $f'(1) = 3$ .
$f''(x) > 0$ for all $x$ , $f'(0) = 1$ , and $f(x) \leq 100$ for all $x > 0$ .
$f''(x) > 0$ for all $x$ , $f'(0) = 1$ , and $f(x) \leq 100$ for all $x < 0$ .

There can be no function meeting all of these conditions since $f''(x) > 0$ implies $f'(x)$ is an increasing function (since its derivative, $f''$ , is positive). But then $f'(0) > f'(1)$ contradicts that $f'(x)$ is increasing.
Let $f(x) = x^2 + x$ . Then
$\begin{align*} f'(x) = 2x + 1 && \implies f'(0) &= 1, \\ && \implies f'(1) &= 3 \end{align*}$

Furthermore, $f''(x) = 2 > 0$ for all $x$ .
There can be no function meeting all of these conditions. Again, $f''(x) > 0$ for all $x$ implies that $f'(x)$ is increasing for all $x$ . Therefore, $f'(0) = 1$ implies $f'(x) > 1$ for all $x > 0$ . Then, by the mean-value theorem, we know that for any $b > 0$ there exists some $c \in (0,b)$ such that
$f(b) - f(0) = f'(c) (b-0) \quad \implies \quad f(b) > b + f(0) \qquad (\text{since } f'(c) > 1).$

Now, choose $b > 100 - f(0)$ . Then, $f(b) > 100$ , contradicting that $f(x) \leq 100$ for all $x$ .
We’ll define $f$ piecewise as follows
$f(x) = \begin{dcases} 1 + x + x^2 & \text{if } x \geq 0 \\ \frac{1}{1-x} & \text{if } x < 0. \end{dcases}$

Then, we can take the derivative of each piece (and see that they are equal, so the derivative is defined at $x = 0$ )

$f'(x) = \begin{dcases} 1 + 2x &\text{if } x \geq 0 \\ \frac{1}{(1-x)^2} & \text{if } x < 0. \end{dcases}$

Taking the derivative again we find

$f''(x) = \begin{dcases} 2 & \text{if } x \geq 0 \\ \frac{2}{(1-x)^3} & \text{if } x < 0. \end{dcases}$

Thus, $f''(x) > 0$ for all $x$ and $f'(0) = 1$ . Furthermore, for $x < 0$ we have

$f(x) = \frac{1}{1-x} \leq 100 \text{ for all } x < 0.$

Prove some inequalities using the mean value theorem

by RoRi

September 28, 2015

Prove the following inequalities using the mean value theorem:

$| \sin x - \sin y| \leq |x - y|$ .
$ny^{n-1} (x-y) \leq x^n - y^n \leq nx^{n-1} (x-y) \qquad \text{if } 0 < y \leq x$ , for $n = 1, 2, 3, \ldots$ .

Proof. Define $f(t) = \sin t$ and $g(t) = t$ . Then $f$ and $g$ are continuous and differentiable everywhere so we may apply the mean value theorem. We obtain
$\begin{align*} &&f'(c) (g(x) - g(y)) &= g'(c) (f(x) - f(y)) & (\text{for some } c \in (x,y))\\ \implies && (\cos c) (x-y) &= \sin x - \sin y \\ \implies && | \cos c | | x-y| &= | \sin x - \sin y| \\ \implies && |x-y| &\geq | \sin x - \sin y|. \end{align*}$

The final step follows since $| \cos c | \leq 1$ for all $c. \qquad \blacksquare$
Proof. Let $f(t) = t^n$ , $g(t) = t$ , then $f'(t) = nt^{n-1}$ and $g'(t) = 1$ . So, by the mean-value theorem we have there exists a $c \in (x,y)$ such that,
$\begin{align*} && f'(c) (g(x) - g(y)) &= g'(c)(f(x) - f(y)) \\ \implies && nc^{n-1}(x-y) &= x^n - y^n. \end{align*}$

But, since $x^{n-1}$ is an increasing function on the positive real axis, and we have $0 < y \leq c \leq x$ we know

$y^{n-1} \leq c^{n-1} \leq x^{n-1}.$

Further, since $(x-y) \geq 0$ and $n$ is positive we can multiply all of the terms in the equality by $n (x-y)$ without reversing inequalities to obtain,

$ny^{n-1}(x-y) \leq nc^{n-1} (x-y) \leq nx^{n-1} (x-y).$

Therefore, substituting $nc^{n-1} (x-y) = x^n - y^n$ from above we obtain the requested inequality:

$ny^{n-1} (x-y) \leq x^n - y^n \leq nx^{n-1} (x-y). \qquad \blacksquare$

Prove properties about the zeros of a polynomial and its derivatives

by RoRi

September 28, 2015

Consider a polynomial $f$ . We say a number $\alpha$ is a zero of multiplicity $m$ if

$f(x) = (x - \alpha)^m g(x),$

where $g(\alpha) \neq 0$ .

Prove that if the polynomial $f$ has $r$ zeros in $[a,b]$ , then its derivative $f'$ has at least $r-1$ zeros in $[a,b]$ . More generally, prove that the $k$ th derivative, $f^{(k)}$ has at least $r-k$ zeros in the interval.
Assume the $k$ th derivative $f^{(k)}$ has exactly $r$ zeros in the interval $[a,b]$ . What can we say about the number of zeros of $f$ in the interval?

Proof. Let $\alpha_1, \ldots, \alpha_k$ denote the $k$ distinct zeros of $f$ in $[a,b]$ and $m_1, \ldots, m_k$ their multiplicities, respectively. Thus, the total number of zeros is given by,
$r = \sum_{i=1}^k m_i.$

By the definition given in the problem, if $\alpha_i$ is a zero of $f$ of multiplicity $m_i$ then

$f(x) = (x - \alpha_i)^{m_i} g(x) \qquad \text{where } g(x) \neq 0.$

Taking the derivative (using the product rule), we have

$\begin{align*} f'(x) &= m_i (x - \alpha_i)^{m_i - 1} g(x) + (x- \alpha_i)^{m_i} g'(x)\\ &= (x - \alpha_i)^{m_i - 1} (m_i g(x) + (x - \alpha_i) g'(x)) \end{align*}$

Thus, again using the definition given in the problem, $\alpha_i$ is a zero of $f'(x)$ of multiplicity $m_i -1$ .
Next, we know from the mean-value theorem for derivatives, that for distinct zeros $\alpha_i$ and $\alpha_j$ of $f$ there exists a number $c \in [\alpha_i, \alpha_j]$ (assuming, without loss of generality, that $\alpha_i < \alpha_j$ ) such that $f'(c) = 0$ . Hence, if $f$ has $k$ distinct zeros, then the mean value theorem guarantees $k-1$ numbers $c$ such that $f'(c) = 0$ . Thus, $f'$ has at least:

$\left(\sum_{i=1}^k (m_i -1)\right) + k - 1 = \left( \sum_{i=1}^k m_i \right) - 1 = r-1 \ \ \text{zeros}.$

By induction then, the $k$ th derivative $f^{(k)}(x)$ has at least $r-k$ zeros.
If the $k$ th derivative $f^{(k)}$ has exactly $r$ zeros in $[a,b]$ , then we can conclude that $f$ has at most $r+k$ zeros in $[a,b]$ .

Prove an alternate expression for the mean-value formula

by RoRi

September 27, 2015

Prove that the expression

$f(x+h) = f(x) + hf'(x+ \theta h) \qquad \text{where } 0 < \theta < 1$

is an equivalent form of the mean-value theorem.
Find the value of $\theta$ in terms of $x$ and $h$ when:

$f(x) = x^2$ ;
$f(x) = x^3$ .

For parts (a) and (b) keep $x$ fixed with $x \neq 0$ and find the limit of $\theta$ as $h$ tends to 0.

Proof. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then by the mean-value theorem we have

$f(b) - f(a) = f'(c) (b-a) \qquad \text{for some } c \in [a,b].$

Letting $a = x$ and $b = x + h$ for some $h > 0$ (since $b > a$ ), we have

$c \in [a,b] \quad \implies \quad c = x + \theta h \qquad \text{for some } \theta \in (0,1).$

(This follows since from our definitions, $h$ is the distance from $b-a$ . Then, since $c$ is somewhere in the interval $[a,b]$ its value must be $a$ plus some portion of the distance to $b$ . This portion is then $\theta$ , which is how we know $0 < \theta < 1$ .) Substituting $x = a$ and $x + h = b$ and $c = x + \theta h$ ,

$\begin{align*} f(b) - f(a) = f'(c) (b-a) && \implies && f(x+h) - f(x) &= f'(x+ \theta h)(x+h-x) \\ && \implies && f(x+h) &= f(x) + h f'(x + \theta h), \end{align*}$

where $0 < \theta < 1$ and $h > 0. \qquad \blacksquare$

Now for parts (a) and (b).

If $f(x) = x^2$ , we have $f'(x) = 2x$ , so,
$\begin{align*} f(x+h) = f(x) + hf'(x+ \theta h) && \implies && (x+h)^2 &= x^2 + 2h (x + \theta h) \\ && \implies && h^2 &= 2 \theta h^2 \\ && \implies && \theta &= \frac{1}{2} \\ && \implies && \lim_{h \to 0} \theta = \frac{1}{2}. \end{align*}$
If $f(x) = x^3$ , we have $f'(x) = 3x^2$ . So,
$\begin{align*} &&f(x+h) &= f(x) + hf'(x+ \theta h) \\ \implies && (x^3 + 3x^2 h + 3xh^2 + h^3) &= x^3 + 3hx^2 + 6 \theta x h^2 + 3 \theta^2 h^3. \\ \implies && 0 &= (3h^3)\theta^2 + (6xh^2) \theta - (3xh^2 + h^3) \\ \implies && 0 &= h \theta^2 + 2x \theta + \left( -x - \frac{h^2}{3} \right) \\ \intertext{Using the quadratic formula to solve for $\theta$ and noting that since $0 < \theta < 1$ only the positive root is possible, we continue computing...} \implies && \theta &= \frac{-2x + \sqrt{4x^2 + 4hx + 4h^2/3}}{2h} \\ \implies && \theta &= \frac{\sqrt{x^2 + xh + h^2/3} - x}{h} \\ \implies && \theta &= \frac{ \left( \sqrt{x^2 + xh + h^2/3} - x \right) \left( \sqrt{x^2 + xh + h^2/3} + x \right)}{h \left( \sqrt{x^2 + hx + h^2/3} + x \right)} \\ \implies && \theta &= \frac{x^2 + xh + \frac{h^2}{3} - x^2}{h \left( \sqrt{x^2 + hx + h^2/3} + x \right)} \\ \implies && \theta &= \frac{x + \frac{h}{3}}{x + \sqrt{x^2 + xh + h^2/3}} \\ \implies && \lim_{h \to 0} \theta &= \frac{1}{2}. \end{align*}$

Show that x^2 = x*sin x + cos x for exactly two real numbers x

by RoRi

September 27, 2015

Consider the equation

$x^2 = x \sin x + \cos x.$

Show that there are two values of $x \in \mathbb{R}$ such that the equation is satisfied.

Proof. Let $f(x) = x^2 - x \sin x - \cos x$ . (We want then to find the zeros of this function since these will be the points that $x^2 = x \sin + \cos x$ .) Then,

$f'(x) = 2x - \sin x - x \cos x + \sin x = x(2 - \cos x).$

Since $2 - \cos x \neq 0$ for any $x$ (since $\cos x \leq 1$ ), we have

$f'(x) = 0 \quad \iff \quad x = 0.$

Then, $f$ is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know $f$ has at most two zeros (if there were three or more, say $x_1, x_2,$ and $x_3$ , then there must be distinct numbers $c_1$ and $c_2$ with $x_1 < c_1 < x_2 < c_2 < x_3$ such that $f'(c_1) = f'(c_2) = 0$ , but we know there is only one $c$ such that $f'(c) = 0$ ).
Furthermore, $f$ has at least two zeros since $f(-\pi) = \pi^2 + 1 > 0$ , $f(0) = -1 < 0$ , and $f(\pi) = \pi^2 + 1 > 0$ . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of $f$ is at most two and at least 2. Hence, the number of zeros must be exactly two $. \qquad \blacksquare$

Explain why the absence of a zero does not violate Rolle’s theorem

by RoRi

September 27, 2015

Consider the function

$f(x) = 1 - x^{\frac{2}{3}}.$

Show that $f(1) = 0$ and $f(-1) = 0$ , but that the derivative $f'(x) \neq 0$ for all $x \in [-1,1]$ . Explain why this does not violate Rolle’s theorem.

Proof. First, we show that $f(1) = 0$ and $f(-1) = 0$ by a direct computation:

$\begin{align*} f(1) = 1 - (1)^{\frac{2}{3}} = 1 - 1 &= 0, \\ f(-1) = 1 - (-1)^{\frac{2}{3}} = 1 - (1^2)^\frac{1}{3} = 1 - 1 &= 0. \end{align*}$

Then, we compute the derivative,

$f(x) = 1 - x^{\frac{2}{3}} \quad \implies \quad f'(x) = -\frac{2}{3} x^{-\frac{1}{3}}.$

To show $f'(x) \neq 0$ for any $x \in [-1,1]$ we consider three cases:

If $x < 0$ then $x^{-\frac{1}{3}} < 0$ implies $f'(x) > 0$ (since $-\frac{2}{3}$ times a negative is positive).
If $x > 0$ then $x^{-\frac{1}{3}} > 0$ implies $f'(x) < 0$ (since $-\frac{2}{3}$ times a positive is then negative).
If $x = 0$ , then $f'(x)$ is undefined (since $x^{-\frac{1}{3}} = \frac{1}{x^{1/3}}$ ).

Thus, $f'(x) \neq 0$ for any $x \in [-1,1]. \qquad \blacksquare$

This is not a violation of Rolle’s theorem since the theorem requires that $f(x)$ be differentiable for all $x$ on the open interval $(-1,1)$ . Since $f'(x)$ is not defined at $x = 0$ , we have $f(x)$ is not differentiable on the whole interval. Hence, Rolle’s theorem does not apply.

Show a function satisfies the hypotheses of the mean value theorem

by RoRi

September 26, 2015

Let

$f(x) = \begin{dcases} \frac{3-x^2}{2} & \text{if } x \leq 1, \\ \frac{1}{x} & \text{if } x \geq 1. \end{dcases}$

Draw the graph of $f$ for $x \in [0,2]$ .
Show that the hypotheses of the mean value theorem are satisfied on $[0,2]$ and find the mean values the theorem provides.

The sketch is as follows:
Since $\frac{3-x^2}{2}$ and $\frac{1}{x}$ are continuous on $[0,1]$ and $[1,2]$ , respectively and are differentiable on $(0,1)$ and $(1,2)$ , the only point at which this piecewise function might be discontinuous or non differentiable is at $x = 1$ .
At $x = 1$ we have

$\lim_{x \to 1^-} \frac{3-x^2}{2} = 1, \qquad \lim_{x \to 1^+} \frac{1}{x} = 1.$

Thus, the left and right-hand limits are equal, so the limit exists and equals the function value. Therefore, $f$ is continuous at $x = 1$ . Finally, we must check that the derivative exists at $x = 1$ . Since the derivative of $\frac{3-x^2}{2} = -x$ and the derivative of $\frac{1}{x} = -\frac{1}{x^2}$ , both are equal to $-1$ at $x = 1$ . Hence, the derivative exists.

Now, we have met the conditions of the mean-value theorem, so we can apply the theorem to conclude,

$\begin{align*} f(2) - f(0) = f'(c)\cdot 2 &\implies \frac{1}{2} - \frac{3}{2} = 2 f'(c) \\ &\implies f'(c) = -\frac{1}{2}. \end{align*}$

Further, we know

$f'(x) = \begin{dcases} -x & \text{if } x < 1 \\ -\frac{1}{x^2} & \text{if } x \geq 1. \end{dcases}$

So,

$f'(c) = -c = -\frac{1}{2} \quad \implies \quad c = \frac{1}{2}$

and

$f'(c) = -\frac{1}{c^2} = -\frac{1}{2} \quad \implies \quad c = \sqrt{2} \qquad (\text{since } -\sqrt{2} \notin [1,2].)$

Prove a property of the integral of the product of continuous functions

by RoRi

September 11, 2015

Let $f$ be a continuous function on the interval $[a,b]$ and assume

$\int_a^b f(x) g(x) \, dx = 0$

for every function $g$ which is continuous on the interval $[a,b]$ . Prove that $f(x) = 0$ for all $x \in [a,b]$ .

Proof. Since $\int_a^b f(x) g(x) \, dx = 0$ must hold for every function $g$ that is continuous on $[a,b]$ , it must hold for $f$ itself (since $f$ is continuous on $[a,b]$ by hypothesis). Therefore we must have,

$\int_a^b f(x) f(x) \, dx = \int_a^b (f(x))^2 \, dx = 0$

However, $(f(x))^2 \geq 0$ for all $x \in [a,b]$ . We know from the previous exercise (Section 3.20, #7) that a non-negative function whose integral is zero on an interval must be zero at every point at which it is continuous. By hypothesis $f$ is continuous at every point of $[a,b]$ ; hence, $f^2$ is also continuous at every point of $[a,b]$ (since the product of continuous functions is continuous). Therefore,

$(f(x))^2 = 0 \qquad \text{for all } x \in [a,b].$

Which implies,

$f(x) = 0 \qquad \text{for all } x \in [a,b]. \qquad \blacksquare$